The ball has two projections of speed: v x v_{x} v x and v y v_{y} v y .
At the initial moment:
v 0 x = v 0 × cos ( α ) v 0 y = v 0 × sin ( α ) v_{0x}=v_{0}\times \cos(\alpha) \newline
v_{0y}=v_{0}\times \sin(\alpha) v 0 x = v 0 × cos ( α ) v 0 y = v 0 × sin ( α )
The x-projection of speed will not change in time.
The y-projection:
v y = v 0 y + a t = v 0 y − g t v_{y}=v_{0y}+at=v_{0y}-gt v y = v 0 y + a t = v 0 y − g t
Take g = 10 g=10 g = 10 :
v x = v 0 x = 12 × cos ( 55 ) ≈ 6.88 v y = 12 × sin ( 55 ) − 10 × 2.16 ≈ − 11.76 v_{x}=v_{0x}=12\times\cos(55)\approx6.88\newline
v_{y}=12\times\sin(55)-10\times2.16\approx-11.76 v x = v 0 x = 12 × cos ( 55 ) ≈ 6.88 v y = 12 × sin ( 55 ) − 10 × 2.16 ≈ − 11.76
So:
v = v x 2 + v y 2 = 47.3344 + 138.2976 ≈ 13.63 m / s v=\sqrt{v_{x}^2+v_{y}^2}=\sqrt{47.3344+138.2976}\approx13.63m/s v = v x 2 + v y 2 = 47.3344 + 138.2976 ≈ 13.63 m / s
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